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Chapter 6: Problem 40

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$x=\frac{1}{4}\left(y^{2}+2 y+33\right)$$

Short Answer

Expert verified
The vertex of the parabola is at \((-1, 8)\), the focus is at \((9, -1)\), and the directrix is at \(x=7\).

Step by step solution

01

Rewrite the Equation in Vertex Form

Rewrite the equation in the form \(x = \frac{1}{4} (y-k)^2 + h\). This can be achieved by completing the square. The given equation is \(x=\frac{1}{4}\left(y^{2}+2 y+33\right)\), which can be rewritten as \(x=\frac{1}{4}\left((y+1)^{2}+32\right)\), so the vertex is \(-1, 8\).
02

Find the Focus

The equation \(x - h = \frac{1}{4p}(y-k)^2\) can be used to find the focus. Here, \(\frac{1}{4} = \frac{1}{4p}\), so \(p = 1\). The focus is then at \(x = h + p = 8 + 1 = 9\). For a parabola opening to the right, the focus is at \((h+p, k)\), so the focus is at \((9, -1)\).
03

Identify the Directrix

The equation of the directrix is \(x = h - p\). So, the directrix of the parabola is at \(x = 8 - 1 = 7\).
04

Sketch the Parabola

With the vertex at \((-1, 8)\), focus at \((9, -1)\), and directrix at \(x=7\), sketch the parabola. The parabola will open towards the right as the equation is of the form \(x = ay^2\). mark these values on an xy-plane, draw the directrix and sketch the parabola opening towards the right, paying attention to maintain the symmetry about the line \(y = k\).

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